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Difference between revisions of "2006 AMC 10B Problems/Problem 14"

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== Solution ==
 
== Solution ==
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In a [[quadratic equation]] in the form <math> x^2 + bx + c = 0 </math>, the product of the roots is <math>c</math>
+
In a [[quadratic equation]] in the form <math> x^2 + bx + c = 0 </math>, the product of the roots is <math>c</math>.
  
 
Using this property:
 
Using this property:
  
−
<math>ab=2</math>
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<math>ab=2</math>.
  
 
<math> q = (a+\frac{1}{b})\cdot(b+\frac{1}{a}) = (\frac{ab+1}{b})\cdot(\frac{ab+1}{a}) = \frac{(ab+1)^2}{ab} = \frac{(2+1)^2}{2} = \frac{9}{2} \Rightarrow D </math>
 
<math> q = (a+\frac{1}{b})\cdot(b+\frac{1}{a}) = (\frac{ab+1}{b})\cdot(\frac{ab+1}{a}) = \frac{(ab+1)^2}{ab} = \frac{(2+1)^2}{2} = \frac{9}{2} \Rightarrow D </math>

Revision as of 13:26, 18 July 2006

Problem

Let $a$ and $b$ be the roots of the equation $x^2-mx+2=0$. Suppose that $a+(1/b)$ and $b+(1/a)$ are the roots of the equation $x^2-px+q=0$. What is $q$?

$\mathrm{(A) \ } \frac{5}{2}\qquad \mathrm{(B) \ } \frac{7}{2}\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } \frac{9}{2}\qquad \mathrm{(E) \ } 8$

Solution

In a quadratic equation in the form $x^2 + bx + c = 0$, the product of the roots is $c$.

Using this property:

$ab=2$.

$q = (a+\frac{1}{b})\cdot(b+\frac{1}{a}) = (\frac{ab+1}{b})\cdot(\frac{ab+1}{a}) = \frac{(ab+1)^2}{ab} = \frac{(2+1)^2}{2} = \frac{9}{2} \Rightarrow D$

See Also

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